Problem: On the way to school, Karen became fixated on the puzzle game on her phone!

<image>

The game is played as follows. In each level, you have a grid with n rows and m columns. Each cell originally contains the number 0.

One move consists of choosing one row or column, and adding 1 to all of the cells in that row or column.

To win the level, after all the moves, the number in the cell at the i-th row and j-th column should be equal to gi, j.

Karen is stuck on one level, and wants to know a way to beat this level using the minimum number of moves. Please, help her with this task!

Input

The first line of input contains two integers, n and m (1 ≤ n, m ≤ 100), the number of rows and the number of columns in the grid, respectively.

The next n lines each contain m integers. In particular, the j-th integer in the i-th of these rows contains gi, j (0 ≤ gi, j ≤ 500).

Output

If there is an error and it is actually not possible to beat the level, output a single integer -1.

Otherwise, on the first line, output a single integer k, the minimum number of moves necessary to beat the level.

The next k lines should each contain one of the following, describing the moves in the order they must be done:

  * row x, (1 ≤ x ≤ n) describing a move of the form "choose the x-th row". 
  * col x, (1 ≤ x ≤ m) describing a move of the form "choose the x-th column". 



If there are multiple optimal solutions, output any one of them.

Examples

Input

3 5
2 2 2 3 2
0 0 0 1 0
1 1 1 2 1


Output

4
row 1
row 1
col 4
row 3


Input

3 3
0 0 0
0 1 0
0 0 0


Output

-1


Input

3 3
1 1 1
1 1 1
1 1 1


Output

3
row 1
row 2
row 3

Note

In the first test case, Karen has a grid with 3 rows and 5 columns. She can perform the following 4 moves to beat the level:

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In the second test case, Karen has a grid with 3 rows and 3 columns. It is clear that it is impossible to beat the level; performing any move will create three 1s on the grid, but it is required to only have one 1 in the center.

In the third test case, Karen has a grid with 3 rows and 3 columns. She can perform the following 3 moves to beat the level:

<image>

Note that this is not the only solution; another solution, among others, is col 1, col 2, col 3.

Solution (in JAVA):

Solution:
import java.io.*;
import java.util.*;


import static java.lang.Math.*;
import static java.util.Collections.*;
import static java.util.Arrays.*;




public class ProblemC {

	static class Problem {

		Scanner reader;
		PrintWriter writer;

		Problem() {
			reader = new Scanner(System.in);
			writer = new PrintWriter(System.out);
		}

		Problem(File inputFile, File outputFile) throws IOException {
			reader = new Scanner(inputFile);
			writer = new PrintWriter(outputFile);
        }
        
        void runSolution() {
            solution();
            writer.close();
        }

        void solution() {
            int n = reader.nextInt();
            int m = reader.nextInt();
            int[][] g1 = new int[n][m];
            int[][] g2 = new int[n][m];
            int[] rowCnt1 = new int[n];
            int[] rowCnt2 = new int[n];
            int[] colCnt1 = new int[m];
            int[] colCnt2 = new int[m];
            int k1 = 0;
            int k2 = 0;
            boolean isPossible = true;

            for (int i = 0; i < n; i++) {
                for (int j = 0; j < m; j++) {
                    g1[i][j] = g2[i][j] = reader.nextInt();
                }
            }

            for (int i = 0; i < n; i++) {
                int minimal = 500;
                for (int j = 0; j < m; j++) {
                    minimal = min(minimal, g1[i][j]);
                }
                for (int j = 0; j < m; j++) {
                    g1[i][j] -= minimal;
                }
                k1 += minimal;
                rowCnt1[i] += minimal;
            }
            for (int i = 0; i < m; i++) {
                int minimal = 500;
                for (int j = 0; j < n; j++) {
                    minimal = min(minimal, g2[j][i]);
                }
                for (int j = 0; j < n; j++) {
                    g2[j][i] -= minimal;
                }
                k2 += minimal;
                colCnt2[i] += minimal;
            }

            for (int i = 0; i < m; i++) {
                int minimal = 500;
                for (int j = 0; j < n; j++) {
                    minimal = min(minimal, g1[j][i]);
                }
                for (int j = 0; j < n; j++) {
                    g1[j][i] -= minimal;
                    isPossible = isPossible && (g1[j][i] == 0);
                }
                k1 += minimal;
                colCnt1[i] += minimal;
            }
            for (int i = 0; i < n; i++) {
                int minimal = 500;
                for (int j = 0; j < m; j++) {
                    minimal = min(minimal, g2[i][j]);
                }
                k2 += minimal;
                rowCnt2[i] += minimal;
            }

            if (!isPossible) {
                writer.println(-1);
            } else if (k1 < k2) {
                writer.println(k1);
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < rowCnt1[i]; j++) {
                        writer.println("row " + (i + 1));
                    }
                }
                for (int i = 0; i < m; i++) {
                    for (int j = 0; j < colCnt1[i]; j++) {
                        writer.println("col " + (i + 1));
                    }
                }
            } else {
                writer.println(k2);
                for (int i = 0; i < n; i++) {
                    for (int j = 0; j < rowCnt2[i]; j++) {
                        writer.println("row " + (i + 1));
                    }
                }
                for (int i = 0; i < m; i++) {
                    for (int j = 0; j < colCnt2[i]; j++) {
                        writer.println("col " + (i + 1));
                    }
                }
            }
        }

    }

    public static void main(String args[]) throws IOException {
        Problem problemC = new Problem();
        //Problem problemC = new Problem(new File("input.txt"), new File("output.txt"));
        problemC.runSolution();
    }

}